# Graham’s Law of Diffusion | Explanation and Application | Chemistry Grade XI

### Graham’s Law of Diffusion # Graham’s Law of Diffusion | Explanation and Application

GRAHAM’S LAW OF DIFFUSION
The spontaneous natural process of intermixing of non-reacting gaseous form homogeneous mixture irrespective of gravitational force and without any external agent is called diffusion. Similarly, the process of escaping gas which is kept under pressure in an enclosed vessel through a small hole(orifice) is called effusion.

Rate of diffusion is defined as the volume of gas diffused per unit time which depends on density of that gas. Thomas Graham gave the relation between rate of diffusion and density of that gas which is known as Graham’s law of diffusion.

Graham’s law of diffusion states that "under the similar condition of temperature and pressure, rate of diffusion is inversely proportional to the square root of its density".
i.e. r is proportional to 1/1√d
where, r = rate of diffusion
d = density of gas

EXPLANATION
Let ‘r1’ and ‘r2’be the rate of diffusion of the gases having density ‘d1’ and ‘d2’. So,
r1 α 1/ √d1  -----------(i)
r2 α 1/√d2 -----------(ii)

Dividing equation (i) and (ii)
r1/ r2= √d2/√d---------(iii)

Molecular mass(M) = 2 * vapour density(d)
Accordingly,
d= M1 / 2
d= M2­ / 2

Put the values of d1 and d2  in equation (iii)
r1/r2 =  √d2 / √d1 = √(M2/2)/√(M1/2) = √M2/√M-----(iv)

Again, Rate of diffusion= volume of gas diffused / time taken
r= V / t
So, r1= V1/ t1
r2 = V2/t2

Putting the value of r1 and r2  are in eqn (iv)
r1/r2 = (V1 /t1) / (V2/ t2) = √d/ √d= √M2/√M----------(v)

If equal volume of gases are diffused then V1 = V2 So equation becomes:
r1/r2 = t/ t1 = √d2 / √d2 = √M/√M1

APPLICATION OF GRAHAM’S LAW OF DIFFUSION

1. It is used to measure the molecular mass of gas.
2. It is used to determine the vapour density of the gas.
3. It is used to separate isotopes.
4. It is used to separate the gases mixture.
5. It is used to measure the rate of diffusion.

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